Why Stochastic Volatility Can Depend on Asset Prices Without Correlated Shocks
Summary
The discussion distinguishes zero correlation between the Brownian shocks in a stochastic volatility model from independence between the asset price and volatility processes. The accepted explanation frames independence in terms of joint and marginal distributions: if two variables are independent, observing one gives no information about the other. This is a property of their joint behavior, not a claim about causal influence.
The answer argues that persistent volatility generally makes asset prices more dispersed. Thus, observing an unusually high or low price can provide information about volatility, even when the instantaneous shocks have zero correlation. It offers intuition rather than a formal proof: whether independence holds depends on the specified volatility dynamics, which are left unspecified. The note also motivates why conditional expected variance given a terminal price may vary with that price, so zero shock correlation alone does not justify a flat local volatility function.
Key ideas
- Zero correlation between driving shocks does not establish independence between the processes they drive.
- Independence concerns whether information about one variable changes beliefs about the other.
- Persistent volatility can make extreme asset prices more likely when volatility is high.
- The dependence argument is intuitive and depends on the specific volatility dynamics.
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# Independence vs correlation in stochastic vol models
# Independence vs correlation in stochastic vol models
I am struggling a bit with some basic stuff lately:
Consider a SV model \begin{align} dS_t &= \sigma_t S_t dW_t \\ d\sigma_t &= b(\sigma_t,t) dZ_t \end{align} with $dW_t dZ_t = 0$.
I know that zero correlation does not imply independence, and in fact $S_t$ is clearly not independent of $\sigma_t$.
However, I cannot see from the above SDEs how $\sigma_t$ can depend on $S_t$, in fact I think it doesn't.
But if $\sigma_t$ were independent of $S_t$, then the local volatility function $$ LV(K,T) := E_t [ \sigma^2_T | S_T = K] = E_t [ \sigma^2_T] $$ would not depend on $K$. But this would imply a flat local vol function which doesn't make sense.
What is wrong in my reasoning?
## Answer by fes (score 1, accepted)
https://quant.stackexchange.com/a/71267
Consider the random variables $S_t$ and $\sigma_t$. Let their marginal cumulative distribution functions be $F_{S,t}$ and $F_{\sigma,t}$. The variables are said to be independent if their joint distribution function satisfies $F_{S,\sigma,t}=F_{S,t}F_{\sigma,t}$. Independence is naturally symmetric. The best way to think about it intuitively is that knowing the value of one random variable gives no information about the other variable. Note that dependence between the random variables might not be in any sense causal.
Without specifying $b$ in your example it is difficult to formally prove independence/dependence. However, when $\sigma_t$ is persistent (autocorrelated), since $S_t=S_0+\int_0^{t}\sigma_sS_sdW_s$ you would expect high values of $\sigma_t$ to increase the dispersion of $S_t$, i.e. make observing extreme values of $S_t$ more likely. Alternatively, a high value of $\sigma_t$ is more likely when $S_t$ takes a value in the tails. Therefore you would not expect these two processes to generally be independent.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.